We know that $0\le\dfrac{\sin^2(n)}{n^4} \le\dfrac{1}{n^4}$ for any $n\ge 1$. Considering this fact, what does the direct comparison test say about $\sum\limits_{n=1}^{\infty }\dfrac{\sin^2(n)}{n^4}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A The series converges. (Choice B) B The series diverges. (Choice C) C The test is inconclusive.
$\sum\limits_{n=1}^{\infty }~{\frac{1}{{{n}^{4}}}}~~$ is a $~p$ -series with $~p=4~$, so it converges. Our given series is term-by-term less than a convergent series, so it converges as well.